Question

 

1) 25.00 mL of acetic acid (Ka = 1.8 x 10-5) is titrated with a 0.09991M NaOH. It takes 49.50 mL of base to fully neutralize the acid and reach equivalence.

What is the concentration of the acid sample?

What is the initial pH?

What is the pH after 20.00 mL of base has been added?

What is the pH at equivalence ?

What is the pH after 60.00 mL of base has been added?

What indicator should be used for this titration?

What is the pKa of the weak acid?

2) Two students want to prepare 2.0 L of a buffer solution which must be buffered at a pH of 4.75. Student A wants to use acetic acid (CH3COOH, Ka = 1.8 x 10-5) and sodium acetate to prepare the buffer solution, while student B wants to use formic acid (HCOOH, Ka = 1.8 x 10-4) and sodium formate.

Which student chose the better conjugate acid-base pair to prepare the buffer? Why?

Explain the step-by- step procedure that the students will use to prepare the buffer solution:

Answer 1

part a

for acid base nutralisation reaction equalvalent amount of acid reacts with  equalvalent bases

then V₁S₁ = V₂S₂

V₁ = volume of acid, S₁ = concentration of Acid, V₂ = volume of base, S₂ = concentration of base

In this problem V₁ = 25.00 ml, S₁ = X(M) , V₂ = 49.50 ml S₂ = 0.09991 M

then S₁ = 49.50 0.09991/25 = 0.1978 (M)

hence concentration of the base is 0.1978 (M)

part b

acetic acid is aweak acid, for weak acid pH =  1/2 (Pk_a - logc)

Initial pH= 1/2 (Pk_a - logc)

= 1/2(-logk_a  - log 0.1978)

= 1/2(4.75 +0.703) = 2.72

part c

After Adding 20.00 ml base

[acid] = initial acid concentration - concentration of the base = (25.000.1978) - (20.000.09991) = 4.9675- 1.9982

= 2.9693 molar

and [ salt ] = [ base] = 1.9982 molar

after adding the base before neutralisation

the solution becomes an acidic buffer

using Henderson equation

pH = pK_a  + log [acid]/[salt]

pH = 4.75 + log 2.9693/1.9982

pH= 4.75 + log1.48 = 4.75+0.172 = 4.922

part d

At equivalence point , concentration of the salt = 4.9675

now ph for a sal of weak acid and strong base is

therefore pH = 1/2pKw + 1/2pka +1/2 log c

or pH = 7+ (4.75)/2 + 1/2log 4.9675 = 7 +2.375 + (0.696)/2 = 9.723

hence pH at equivalence point is 9.723